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Lecture 14: Basic Number Theory

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Lecture 14: Basic Number Theory Lecture 14: Basic number theory Nitin Saxena ? IIT Kanpur 1 Euler's totient function φ The case when n is not a prime is slightly more complicated. We can still do modular arithmetic with division if we only consider numbers coprime to n. For n ≥ 2, let us define the set, ∗ Zn := fk j 0 ≤ k < n; gcd(k; n) = 1g : ∗ The cardinality of this set is known as Euler's totient function φ(n), i.e., φ(n) = jZnj. Also, define φ(1) = 1. Exercise 1. What are φ(5); φ(10); φ(19) ? Exercise 2. Show that φ(n) = 1 iff n 2 [2]. Show that φ(n) = n − 1 iff n is prime. Clearly, for a prime p, φ(p) = p − 1. What about a prime power n = pk? There are pk−1 numbers less than n which are NOT coprime to n (Why?). This implies φ(pk) = pk − pk−1. How about a general number n? We can actually show that φ(n) is an almost multiplicative function. In the context of number theory, it means, Theorem 1 (Multiplicative). If m and n are coprime to each other, then φ(m · n) = φ(m) · φ(n) . ∗ ∗ ∗ ∗ ∗ Proof. Define S := Zm × Zn = f(a; b): a 2 Zm; b 2 Zng. We will show a bijection between Zmn and ∗ ∗ ∗ ∗ ∗ S = Zm × Zn. Then, the theorem follows from the observation that φ(mn) = jZmnj = jSj = jZmjjZnj = φ(m)φ(n). ∗ The bijection : S ! Zmn is given by the map :(a; b) 7! bm + an mod mn. We need to prove that is a bijection. That amounts to proving these three things. ∗ ∗ ∗ { The mapping is valid, i.e., if a 2 Zm and b 2 Zn then bm + an 2 Zmn. This follows from the fact that bm is coprime to n implies bm + an is coprime to n. Similarly bm + an is coprime to m. So bm + an is coprime to mn (and we use its residue representative in [mn − 1]). { Mapping is injective (one to one). Why? If bm + an = b0m + a0n mod mn implies (b − b0)m + (a − a0)n = 0 mod mn. The latter implies, using coprimality of m; n, that nj(b − b0) and mj(a − a0). Thus, (a; b) = (a0; b0) in S. { Mapping is surjective (onto). Why? ∗ −1 ∗ Consider t 2 Zmn. Compute k := tm mod n. (Note: k 2 Zn.) Since t = km mod n we can write 0 0 0 ∗ t = km + `n. If need be, reduce ` to ` mod m. This achieves both t = km + ` n mod mn and ` 2 Zm. These three properties of finish the proof. Exercise 3. Find numbers m; n such that φ(mn) 6= φ(m)φ(n). ? Edited from Rajat Mittal's notes. Fundamental theorem of arithmetic implies that we can express any number as a product of prime powers. By using Thm 1, we can calculate φ(mn), when φ(m) and φ(n) are given to us (m and n are coprime). k1 k2 k` Theorem 2. If n = p1 p2 ··· p` is a natural number. Then, 1 1 1 φ(n) = n 1 − 1 − ··· 1 − : p1 p2 p` Exercise 4. Prove the above theorem using the argument above. 2 Inclusion-Exclusion vs. M¨obiusInversion There is another way to look at Thm. 2. We are interested in finding out the number of elements between 0 k1 k2 k` and n−1 which do not share a factor with n = p1 p2 ··· p` . Let us consider all the elements f0; 1; ··· ; n−1g. Define Ai to be the set of elements which are divisible by pi. For any I ⊆ [`], define AI to be the set of elements which are divisible by all pi where i 2 I. You can see that we are interested in the event when none of the pi's, where i 2 [`], divide an element. This is a straightforward application of inclusion-exclusion, X jIj φ(n) = (−1) · jAI j : I⊆[`] n Notice that the number of elements which are divisible by p1p2 ··· pj is just . This gives us, p1p2···pj n jAI j = Q : i2I pi So, X n φ(n) = (−1)jIj : (1) Πi2I pi I⊆[`] Exercise 5. Prove that the above expression is the same as the one in Thm. 2. In Eqn. 1, the sum is taken over all square-free (i.e. of the form p1p2 ··· pi with distinct primes) divisors of n. Define a function, µ(k), 8 < 1; if k = 1 µ(k) := 0; if a2 j k for some a ≥ 2 : (−1)r; if k is square-free with r primes. This function µ(k) is called the M¨obiusfunction. Then Eqn. 1 can be rewritten as, X n φ(n) = µ(d) · : d djn P Exercise 6. For an integer n ≥ 2 show that, djn µ(d) = 0 . ] ` [ 2 i ] ` [ 2 i n j d i i i i )=0. = )) p ( µ + (1 = )) p ( µ + ··· + ) p ( µ + ) p ( µ + (1 = ) d ( µ that deduce Q Q P 2 i k .Ti a eue to used be can This ). ab ( µ ) = b ( µ · ) a ( µ , b a; coprime for i.e. function, multiplicative a ) is k ( µ M¨obiusfunction is really useful in number theory, and combinatorics. One of the main reasons is the \inversion property" (for special functions f). Theorem 3 (M¨obiusinversion). Let f and g be functions defined on natural numbers. Then, X X n f(n) = g(d) implies g(n) = µ(d)f : d djn djn 2 Proof. Let us look at RHS of the expression for g(n): X n X X µ(d)f = µ(d) g(c) d djn djn cjn=d 0 1 X X = g(c)@ µ(d)A cjn djn=c = g(n)µ(1) = g(n) : P The third equality follows from the fact that djn µ(d) is 0 for n ≥ 2 (is 1 for n = 1). The second equality is sum-swapping. Exercise 7. Prove the second equality by considering the pairs (c; d) s.t. d j n and c j n=d. Functions f and g are called M¨obiustransforms of each other. Eg. n and φ(n) are M¨obiustransforms of each other! Exercise 8. Finite fields are routinely used in computer science. Read up on how to use M¨obiusinversion to count the number of irreducible polynomials, of degree d, over a finite field. References 1. N. L. Biggs. Discrete Mathematics. Oxford University Press, 2003. 2. P. J. Cameron. Combinatorics: Topics, Techniques and Algorithms. Cambridge University Press, 1994. 3. K. H. Rosen. Discrete Mathematics and Its Applications. McGraw-Hill, 1999. 3.
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